Question: Is ${425295}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {425295}= &&{4}\cdot100000+ \\&&{2}\cdot10000+ \\&&{5}\cdot1000+ \\&&{2}\cdot100+ \\&&{9}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {425295}= &&{4}(99999+1)+ \\&&{2}(9999+1)+ \\&&{5}(999+1)+ \\&&{2}(99+1)+ \\&&{9}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {425295}= &&\gray{4\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {4}+{2}+{5}+{2}+{9}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${425295}$ is divisible by $3$ if ${ 4}+{2}+{5}+{2}+{9}+{5}$ is divisible by $3$ Add the digits of ${425295}$ $ {4}+{2}+{5}+{2}+{9}+{5} = {27} $ If ${27}$ is divisible by $3$ , then ${425295}$ must also be divisible by $3$ ${27}$ is divisible by $3$, therefore ${425295}$ must also be divisible by $3$.